$f(x) = \dfrac{ \sqrt{ x - 5 } }{ x^2 + 7 x + 12 }$ What is the domain of the real-valued function $f(x)$ ?
Solution: $f(x) = \dfrac{ \sqrt{ x - 5 } }{ x^2 + 7 x + 12 } = \dfrac{ \sqrt{ x - 5 } }{ ( x + 4 )( x + 3 ) }$ First, we need to consider that $f(x)$ is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero. So $x - 5 \geq 0$ , which means $x \geq 5$ Next, we also need to consider that $f(x)$ is undefined anywhere where the denominator is zero. So $x \neq -4$ and $x \neq -3$ However, these last two restrictions are irrelevant since $5 > -4$ and $5 > -3$ and so $x \geq 5$ will ensure that $x \neq -4$ and $x \neq -3$ Combining these restrictions, then, leaves us with simply $x \geq 5$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \geq5\, \}$.